The line with equation $y = x$ is an axis of symmetry of the curve with equation
\[y = \frac{px + q}{rx + s},\]where $p,$ $q,$ $r,$ $s$ are all nonzero.  Which of the following statements must hold?

(A) $p + q = 0$

(B) $p + r = 0$

(C) $p + s = 0$

(D) $q + r = 0$

(E) $q + s = 0$

(F) $r + s = 0$
Explanation: Since $y = x$ is an axis of symmetry, if point $(a,b)$ lies on the graph, then so does $(b,a).$  Thus, the equation of the graph can also be written as
\[x = \frac{py + q}{ry + s}.\]Substituting $y = \frac{px + q}{rx + s},$ we get
\[x = \frac{p \cdot \frac{px + q}{rx + s} + q}{r \cdot \frac{px + q}{rx + s} + s} = \frac{p(px + q) + q(rx + s)}{r(px + q) + s(rx + s)}.\]Cross-multiplying, we get
\[x[r(px + q) + s(rx + s)] = p(px + q) + q(rx + s).\]Expanding, we get
\[(pr + rs) x^2 + (s^2 - p^2) x - (pq + qs) = 0.\]We can take out a factor of $p + s$:
\[(p + s)(rx^2 + (s - p) x - q) = 0.\]This equation must holds for all $x.$  Since $r \neq 0,$ the quadratic $rx^2 + (s - p) x - q$ cannot be 0 for all $x,$ so we must have $p + s = 0.$  The correct statement is $\boxed{\text{(C)}}.$